SETS OF BINOMIAL COEFFICIENTS WITH E^UAL PRODUCTS
CALVIN T. LONG
Washington State University, Pullman, Washington 99163
and University of British Columbia, Vancouver, B.C., Canada
and
VERNER E. HOGGATT, JR.
San Jose State University, San Jose, California 95192
1.
INTRODUCTION
In [2] , Hoggatt and Hansell show that the product of the six binomial coefficients s u r rounding any particular entry in P a s c a l ' s triangle is an integral square.
They also observe
that the two products of the alternate triads of these six numbers a r e equal.
Quite r e m a r k -
ably, Gould conjectured and Hillman and Hoggatt [l] have now proved that the two greatest
common divisors of the numbers in the above-mentioned triads are also equal though their
l e a s t common multiples a r e , in general, not equal.
Hillman and Hoggatt also generalize the
greatest common divisor property to more general a r r a y s .
The integral square property was further investigated by Moore [4] , who showed that
the result is true for any regular hexagon of binomial coefficients if the number of entries
p e r side is even, and by the present author [ 3 ] , who generalized the e a r l i e r results to nonregular hexagons, octagons, and other a r r a y s of binomial coefficients whose products are
squares.
In the present paper, we generalize the equal product property of Hoggatt and Hansell
along the lines of [3] and also make some observations and conjectures regarding a generalized greatest common divisor property.
It will suit our purpose to represent P a s c a l ' s triangle (or, more precisely, a portion of
it) by a lattice of dots as in Fig. 1. We will have occasion to refer to various polygonal figu r e s and when we do, unless expressly stated to the contrary, we shall always mean a simple
closed polygonal curve whose vertices are lattice points. Occasionally, it will be convenient
to represent a small portion of P a s c a l ' s triangle by letter arranged in the proper position.
Fig. 1
Fig. 2
71
72
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
2.
[Feb.
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
As in [3] , we begin by deriving a fundamental lemma which is basic to all of the other
results of this section.
Lemma 1. Consider two parallelograms of binomial coefficients oriented as in Fig. 2
and with corner coefficients
a, b, c, d and e , f, g, h a s indicated.
Then the products
acfh and bdeg are equal.
Proof. For suitable integers m, n, r , s , and t, the binomial coefficients in question
may be represented in the form
+ B
* - ( \
c
_/
m+r \
In + r + tJ'
) '
b
=( n ) '
m
f = (
\
yn + r + t y '
c
= (n+\r)>
* = (mn\\+r)'
g
=(
m+s
\
^n + s + r + ty'
,
=
/ m + s + r \
\n + a + r + t J '
Thus, the desired products a r e
a c fh
=
<m + s)'.
(m + r)t
n! (m - n + sjl (n + r)! (m - n)!
ml
_
(m + s + r)l
(n + r + t)! (m - n - r - t)l ' (n + s + r + t)! (m - n - t)!
and
, ,
DQeg
_
m!
(m + s + r)!
n! (m - n)! " W+ r)(m - n + s)!
(m + r)
(m + s)t
(n + r + t)! (m - n - t)t ' (n + r + s + t)! (m - n - r - t)!
and these a r e clearly equal as claimed.
As a first consequence of Lemma 1, we obtain the equal product result of Hoggatt and
Hansell.
Theorem 2. Let a, b, c , d, e , f, and g denote binomial coefficients as in the a r r a y
a
f
b
g
e
c .
d
Then aec = fbd.
Proof.
Therefore,
The parallelograms a, f, e , g and b, g, d, c are oriented as in Lemma 1.
fgbd = aegc and this implies the desired result.
By essentially the same argument, we obtain the following more general statement about
products of coefficients at the vertices of hexagons in P a s c a l ' s triangle.
Theorem 3. Let m > 1 and n > 1 be integers and let H be a convex hexagon whose
sides lie on the horizontal rows and main diagonals of P a s c a l ' s triangle.
Let the number of
1974]
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
coefficients on the respective sides of H be m, n, m , n, m,
a, b, c, d, e,
73
and n in that o r d e r , and let
and f be the coefficients in cyclic order at the vertices of H.
Then
ace =
bdf.
Proof.
Without loss in generality, we may take
m
to be the number of coefficients
along the bottom side of H. If we consider two m-by-n parallelograms with a common v e r tex and with corner coefficients
a, b, c, d, e, f,
and g as in Fig. 3, then, again by Lemma
1, fgbd = aegc and this implies the equality claimed.
Fig.
3
Now, as in [3] , let us call the hexagons of Hoggatt and Hansell fundamental hexagons
and say that a polygonal figure P on P a s c a l ' s triangle is tiled with fundamental hexagons if
P is "covered" by a set F of fundamental hexagons F in such a way that
i.
ii.
iii.
The vertices of each F in F a r e coefficients in P or in the interior of P.
Each boundary coefficient of P is a vertex of precisely one F in F,
and
Each interior coefficient of P Is interior to some F in F o r is a vertex shaped
by precisely two elements of
F.
We can then prove the following result.
Theorem 4.
Let P
be a polygonal figure on P a s c a l ' s triangle with boundary coef-
ficients a1? a 2 , • • • , a n in order around P . If P
can be tiled by fundamental hexagons,
then n = 2s for some s ^- 3 and
s
"
s
E
Proof.
Suppose that P
a
2i-1 = "
1=1
can be tiled with
2i •
1=1
r
fundamental hexagons.
The proof p r o -
ceeds by induction on r. Clearly the least value of r is 1 which occurs only in the Case Of
the fundamental hexagon itself.
In this c a s e , n = 6 and the result is true by Theorem 2.
Now suppose that the result is true for any polygon that can be tiled with fewer than k fundamental hexagons where k > 1 is fixed and let P
damental hexagons.
be a polygon that can be tiled with k fun-
Let H be one of the hexagons which tiles P
boundary point of P . We distinguish five cases.
and contains at least one
74
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
Case 1. H contains just one boundary point of P .
may let a
J
a'
[Feb.
Without loss in generality, we
be the boundary point of P which is in H. Let af , a' ,.., aT 1 0 , a' ini
J F
n
n
n
n+1
n+2
n+3
. denote the other five boundary points of H in o r d e r around H. Let P
and
be the poly-
gon obtained from PT by deleting H. Then the boundary points of P
a r e a 1} a 2 , • • • ,
a .,, af , a'
, a l 0 , a' o 5 and a' , ,. Thus, m = n + 4. Also, since P
can be
n-1
n
n+1
n+2
n+3
n+4
m
tiled by k - 1 fundamental hexagons, n + 4 = m = 2t for some t and
a..1 a63 • • • a - a? ,., a' 0 = za 2 4a 4 • • • a 0 a? af , 0 a f , , .
n - 1 n+1 n+3
n-2 n n+2 n+4
But, since H is a fundamental hexagon, it follows that
a
nan+lan+3 "
a
nan+2an+4
and this clearly implies that
s
s
."
a
2i-i
=
1=1
a
"
2i
1=1
since n = 2t - 4 = 2s. This completes the proof for Case 1.
Cases 2-4. In these c a s e s , respectively,
H contains 2, 3, 4, or 5 boundary points of
P . We omit the proofs of these cases since they essentially duplicate the proof of Case 1,
This completes the proof.
With Theorem 4 and Lemma 1 as our principal tools we a r e now able to give several
quick results.
Theorem 5. Let H
be a convex hexagon with an even number of coefficients per side,
with sides oriented along the horizontal rows and main diagonals of P a s c a l ' s triangle, and with
boundary coefficients
a l 5 a 2 , • • • , a n in order around H . Then n = 2s for some
s ^- 3
and
s
n
s
n
a
2i-i =
1=1
Proof.
a
n
2i •
1=1
This is an immediate consequence of Theorem 4 since H
can be tiled by fun-
damental hexagons as shown in Theorem 5 of [3].
Theorem 6. Let K
be any convex octagon with sides oriented along the horizontal and
vertical rows and main diagonals of P a s c a l ' s triangle and with boundary coefficients
a l9 a 2 ,
• • • , a„n in o r d e r around K . Let the number of coefficients on the various sides of K be
n
n
2r, 2s, t, 2u, 2v, t, and 2s as indicated in Fig. 4. Then n = 2h for some h > 4 and
s
"
1=1
a
2i-l =
s
"
1=1
a
2i
1974]
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
75
2v
2s
2s
2r
Fig. 5
Fig. 4
Proof.
The proof is the same as for Theorem 5 and will be omitted.
We observe that the convexity conditions in both Theorems 3 and 5 a r e n e c e s s a r y since
neither result is true for the hexagon of Fig. 5. Also, it is easy to find examples of convex
hexagons where the results of Theorems 3 and 5 do not hold if the conditions on the number of
elements per side are not met.
In fact, we conjecture that the conditions in these theorems
a r e both necessary and sufficient.
On the other hand, the convexity condition of Theorem 6
is not n e c e s s a r y since the result holds for the octagon of Fig. 6 which is clearly not convex.
We make no conjecture regarding n e c e s s a r y and sufficient conditions for the. result of Theor e m 6 to hold for octagons in general, or indeed, for hexagons whose sides may not lie along
the horizontal rows and main diagonals of P a s c a l ' s triangle.
We note that the octagon of Fig.
6 cannot be tiled by fundamental hexagons but can be tiled by pairs of properly oriented "fundamental parallelograms" as indicated by the shading in the figure.
Thus, the most general
theorem for these and other polygons will most likely have to be couched in t e r m s of tilings
by sets of pairs of fundamental parallelograms.
<$/ <>/ \</ \ /
,\ / ' \ A-\ / Z \ /\> /. >\
/ W
^?
T5:/
N
./
\
A\ i'\
'vh
Fig. 6
Fig. 7
76
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
[Feb.
3. ADDITIONAL COMMENTS ON EQUAL PRODUCTS
Theorem 3 gives an equal product result for the corner coefficients of hexagons and it
is natural to seek similar results for octagons.
It is easy to find octagons like those in Figs.
4 and 6 for which the equal product property on vertices does not hold.
Nevertheless, it is
possible to find classes of octagons for which the equal product property does hold for the
products of alternate corner coefficients.
Theorem 7.
Let K be a convex octagon formed as in Fig. 7 by adjoining parallelo-
g r a m s with r and s and r and t elements on a side to a parallelogram with r elements
on each side.
If the corner coefficients a r e a l5 a 2 , • • • , a8 as shown, then
a1a3a5a7 = a2a4a6a8 .
Proof.
We have only to observe that a1? a 4 , a 5 , a 8 and a 2 , a 3 , a 6 , a 7 are vertices of
p a i r s of parallelograms oriented as in Lemma 1. The result is then immediate.
Again it is clear that the convexity condition of Theorem 7 is not necessary. The proof,
after all, r e s t s on the presence of the properly oriented p a i r s of parallelograms.
In p r e -
cisely the same way we show that a 1 a3 a 5 a 7 = a2 a4 a6 a 8 for each of the three octagons of
Fig. 8. Note that for K2 the two products a r e not products of alternate vertices around the
octagon.
Clearly the preceding methods can be used to obtain a wide variety of configurations
of binomial coefficients which divide into sets with equal products. As illustrations we give
several examples of polygons (sometimes not closed, simple, or connected) with this property.
a7
a4
a7
a6
a
7
a
6
Fig. 8
4.
THE GREATEST COMMON DIVISOR PROPERTY
As mentioned in Section 1, if the a r r a y
a
c
b
d
f
e
g
r e p r e s e n t s coefficients from P a s c a l ' s triangle, then afe = cbg and Hillman and Hoggatt
1974]
S E T S O F BINOMIAL C O E F F I C I E N T S WITH E Q U A L P R O D U C T S
12
12
a
"
2i-l
1=1
=
a
"
*
2i
1=1
a
2i-l
1=1
=
a
"
1=1
2i
II
ai
a2
at
*
a2
*
*
a17
*
a
a3
*
a18*
*
*
24 /
\l9
*
a5
*
*
*
*
*
*
a10
*
*
a23 4
A7
*
*
a
21
*
a
12
3ft
a9
ia 2 0 I a7
a
22
*
*
a4
*
a
n
io
a9
12
12
"
a
2i-l
1=1
= ."
a
.n^i-l
2i
=
III
1=1
IV
a4
a5
i
a
a
s
*
l6
:
an '
*
a
a12
an
10
10
. n - a*02 i - l
. n1 a20 i - l = . n
" a02i
i=l
T
* 1a
3
11
i=l
i=l
VI
Fig.
9
1
20 « 1
a
l
i
11
.V2i
1=1
1=1
r as
*
~
" a2i
i=l
77
78
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
[Feb.
have shown that (a, f, e) = (c, b, g) where we use parentheses to indicate greatest common divisors.
In view of the preceding results on equal products, one wonders if the g r e a t -
est divisor property also holds in more general settings.
Unfortunately, it is easy to find examples of regular hexagons with sides oriented along
the main diagonals and horizontal rows of P a s c a l ' s triangle where the two alternate triads of
corner coefficients have different greatest common divisors in spite of the fact that theyhave
equal products by Theorem 3. We have such examples for hexagons with 3 , 4 , 5 and 6 c o efficients per side and conjecture that the property only holds in general for the fundamental
hexagons of Hoggatt and Hansell.
Also, we observe that, for the parallelograms of Fig. 2,
the products acfh and bdeg a r e equal but that
(a, c, f, h) is not necessarily equal to
(b, d, e, g). At the same time, we have been unable to find examples of hexagons of the type
of Theorem 5 where the greatest common divisor of the two sets of alternate boundary coefficients are not equal.
Of course, these greatest common divisors are usually equal to one,
but the three regular hexagons with four elements per side whose upper left-hand coefficients
a r e , respectively,;
\S)'
Ce)'
and
(")
have p a i r s of greatest common divisors equal to 13, 13, and
34,
respectively.
We con-
jecture that the greatest common divisors of the two sets of alternate boundary coefficients
for the hexagons of Theorem 5 a r e equal.
This is not t r u e , however, of the octagons of Theorem 6, since, in particular,
((:)•
(••)• ( 0 -
( ( ! ) •
( : ) •
( : ) •
(:))->•
(
:
)
and these are alternate boundary coefficients of such an octagon.
)
-
Of c o u r s e , this makes it
clear that not all polygons that can be tiled with fundamental hexagons have the equal greatest
common divisor property.
At the same time, some figures that cannot be tiled with funda-
mental hexagons appear to have the equal greatest common divisor property.
For example,
this appears to be true of the octagon of Fig. 11 in [3] though we have no proof of this fact.
This leaves the question of the characterization of figures having the equal greatest common
divisor property quite open.
5.
GENERALIZATIONS AND EXTENSIONS
There are an infinitude of other Pascal-like a r r a y s in which the Hexagon Squares p r o p erty holds.
F o r example, the Fibonomial triangle and the generalized Fibonomial triangle.
If, indeed we replace F
by f (x), the property holds and thus for each x integral yields
an infinitude of such a r r a y s .
k
For example, if x = 2,
Pell Number Sequence works.
we get the Pell numbers, or every
1974]
SETS OF BINOMIAL COEFFICIENTS WITH EQUAL PRODUCTS
79
REFERENCES
1. A. P. Hillman and V. E. Hoggatt, J r . , "On Gould f s Hexagon Conjecture, Fibonacci Quarterly, Vol. 10, No. 6 (Dec. 1972), pp. 561-564.
2.
V. E. Hoggatt, J r . , and W. Hansell, "The Hidden Hexagon Squares," FibonacciQuarterlXi Vol. 9, No. 2 (Apr. 1971), pp. 120, 133.
3.
Calvin T. Long, "Arrays of Binomial Coefficients whose Products are S q u a r e s , " Fibonacci Quarterly, Vol. 11, No. 5 (Dec. 1973), pp. 449-456.
4.
Carl L. Moore, "More Hidden Hexagon S q u a r e s , " Fibonacci Quarterly, Vol. 11, No. 5
(Dec. 1973), p. 525.
(Continued from page 46. )
that
F
F
3 n (modulo
8. 3 n - i
8- 3 n--
1
3n+1)
1 + 3 n (modulo 3 n + 1 ) .
-!
Therefore
F
8- 3n--i+x
ss
If x satisfies (*), then either
modulo 3
FF
H+
3n(F
X
x or
+ F _,,) (modulo 3 n + 1 ) .
X
8-3
X+l
+ x or
16°3 ~ + x
will be congruent to m
. Therefore (*) has solutions for a r b i t r a r i l y large n.
Problem 2.
The number N is said to have complete Fibonacci residues if there ex-
ists a solution to the congruence
F
= m (modulo N)
for all integers m. A computer search shows that the only values of N ^ 500 having complete Fibonacci residues are the divisors of
35,
2 2 .5 3 ,
2-3-5 3 , 5-3 4 , or 7-53 .
Determine all N which have complete Fibonacci residues.
Problem 3 is submitted by the undersigned and Leonard Carlitz, Duke University, Durham, North Carolina.
Problem 3. Show that if
(Continued on page 82.)
= em'n9
then